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Grade 12th passElectrostatics

two positive point charges of 12uc and 8uc are 10cm apart. the work done in bringing them 4cm closer is?

Profile image of tushar pal
10 Years agoGrade 12th pass
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6 Answers

Profile image of Sourabh Singh
10 Years ago
Hii

Find out the change in the potential energy of the system and take the difference. You will get the answer.


Best of Luck
Profile image of Adarsh srivastav
8 Years ago
q1=12×10^-6,q2=8×10^-6.
Wkt,potential when the are at R1(10 cm)
V1=kq1q2/R1=86.4×10^-1~=86×10^-1.
V2=kq1q2/R2=216×10^-1.
W.D=V2-V1=(216×10^-1)-(86×10^-1)=130×10^-1
                                                                =13J
Profile image of Purba bandyopadhyay
8 Years ago
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Profile image of Krish Gupta
6 Years ago
 
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Profile image of Krish Gupta
6 Years ago
HiiFind out the change in the potential energy of the system and take the difference. You will get the answer.Best of Luck
 
Profile image of Kushagra Madhukar
6 Years ago
Dear student,
Please find the solution to your problem below.
 
Initial distance between the charges, d1 = 10 cm = 0.1 m
Final distance between the charges, d2 = 10 cm – 4 cm = 6 cm = 0.06 m
Hence, W = 9 x 109 x 12 x 10-6 x 8 x 10-6 (1/0.06 – 1/0.1)
               = 5.76 J
 
Hope it helps.
Thanks and regards,
Kushagra

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